The convolution theorem for fourier series.:$ \widehat{f*g}(x) =2π\hat{g}(x)\cdot\hat{f}(x) $

Question

I saw this theorem in Wikipedia, and didn't see it in other place. I tried to proof it, but failed.

How to prove that for $2\pi-$periodic functions one has The convolution theorem for fourier series.: $$ \widehat{f*g}(x) =2π\hat{g}(x)\cdot\hat{f}(x) $$ Note: this is convolution theorem for fourier series, not transforms

see here

Answer 1

The proof use the same token as in the case of Fourier transform. Indeed, $f$ and $g$ be $2\pi$-periodic functions. But since the function $y\mapsto e^{-iyx}g(y)$ is $2π-$periodic, then using this we have , $$\int_{0}^{2\pi} e^{-ix(t-u)}g(t-u)dt\overset{y=t-u}{=} \color{blue}{\int_{-u}^{2\pi-u} e^{-ixy}g(y)dy =\int_{0}^{2\pi} e^{-ixy}g(y)dy}= 2π\hat{g}(x) $$

Inserting this in the following relation we get

$$ \color{blue}{\widehat{f*g}(x) =} \frac{1}{2\pi}\int_{0}^{2\pi}e^{-ixt}f*g(t)dt = \frac{1}{2\pi}\int_{0}^{2\pi}e^{-ixt}\int_{0}^{2\pi} f(u)g(t-u)dudt\\= \frac{1}{2\pi}\int_{0}^{2\pi}\left(e^{-iux}f(u)\int_{0}^{2\pi} e^{-ix(t-u)}g(t-u)dt\right)du\\= 2π\hat{g}(x) \frac{1}{2\pi}\int_{0}^{2\pi}e^{-iux}f(u)du= \color{red}{2π\widehat{g}(x)\cdot\widehat{f}(x)} $$

we used this An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.